Question

find ((4x^4 + 3x^3 - 12x^2 - x + 6)(4x + 3)^{-1}).

Explanation:

Step1: Rewrite the expression

We know that \((4x + 3)^{-1}=\frac{1}{4x + 3}\), so the expression becomes \(\frac{4x^{4}+3x^{3}-12x^{2}-x + 6}{4x + 3}\). We will use polynomial long - division to simplify this fraction.

Let's divide the polynomial \(4x^{4}+3x^{3}-12x^{2}-x + 6\) by \(4x + 3\).

First term: Divide the leading term of the numerator \(4x^{4}\) by the leading term of the denominator \(4x\). \(\frac{4x^{4}}{4x}=x^{3}\)

Multiply the divisor \(4x + 3\) by \(x^{3}\): \((4x + 3)\times x^{3}=4x^{4}+3x^{3}\)

Subtract this from the numerator: \((4x^{4}+3x^{3}-12x^{2}-x + 6)-(4x^{4}+3x^{3})=-12x^{2}-x + 6\)

Second term: Divide the leading term of the new numerator \(-12x^{2}\) by the leading term of the denominator \(4x\). \(\frac{-12x^{2}}{4x}=-3x\)

Multiply the divisor \(4x + 3\) by \(-3x\): \((4x + 3)\times(-3x)=-12x^{2}-9x\)

Subtract this from the new numerator: \((-12x^{2}-x + 6)-(-12x^{2}-9x)=8x + 6\)

Third term: Divide the leading term of the new numerator \(8x\) by the leading term of the denominator \(4x\). \(\frac{8x}{4x}=2\)

Multiply the divisor \(4x + 3\) by \(2\): \((4x + 3)\times2 = 8x+6\)

Subtract this from the new numerator: \((8x + 6)-(8x + 6)=0\)

So, \(\frac{4x^{4}+3x^{3}-12x^{2}-x + 6}{4x + 3}=x^{3}-3x + 2\)

Answer:

\(x^{3}-3x + 2\)